In an inventory analysis it was concluded that, on the average, demands for a certain item were made 5 times per day. What is the probability that on the next 2 days this item is requested not more than 5 times?

which one is correct?

Explanation: Step 1: The average number of times an item requested per day = 5 The distribution of the number of times this item requested on the next 2 days, x~ Poisson (A = 2* 5 =10) Step 2: The probability mass function ofX is, 610 * (10):: P(X=X)= a}! ,X=0,1,2,... Step 3: The probability that on the next 2 days this item is requested not more than 5 times is, P'XS5)=Z:=OP(X :33) he probability that the next two days have less than 5 requests is 5/10 x 4/9 x 3/8 x 2/7 x 1/6 = 0.0292 Therefore, P(less than 5) = 0.0292 P(demand is greater than 5) = 1 - P(demand is not greater than 5). 0.545 x (1-1/2^4) = 0.0292 On the average, demands for a certain item were made 5 times per day. The probability that on the next 2 days this item is requested not more than 5 times is 0.0292. The probability of not requesting it more than 4 times in a row is 0.999999999 and the chance of not requesting it at all on that particular day is 0.9999986667. Therefore, the probability that an item has been requested less than 5 times in 3 days would be: 1-(0.999999999*0)+(0.*9999986667)*(1-5/365) = 1-(0.914793147*0)+(7.65782608775*7.0999816842) = 0.0292 If an item is requested 5 times in 3 days, then it will have been requested 2 times not more than 5 times in a row. On the other hand, if it is requested 1 time on a given day, then it will not have been requested more than 4 times in a row. From this we can conclude that the average demand of this item has been made 5 times and on the next two days there is a high probability